Rewriting log_6(15x) as log_6(15) + log_6(x) and log_6(5x) as log_6(5) + log_6(x)

[doughishere]

Here is what is confusing me:

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Rewriting the given equation, we have:

. . . . .\(\displaystyle 2\, =\, \dfrac{\log_6(15x)}{\log_6(5x)}\)

. . . . .\(\displaystyle 2\, =\, \dfrac{\log_6(15)\, +\, \log_6(x)}{\log_6(5)\, +\, \log_6(x)}\)

Solving the equation for \(\displaystyle \log_6(x),\)

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My question is : What's the rule that just allows you to rewrite the equation this way? I been working problems all day so maybe its just my mind is fried.

TIA.

 

[tkhunny]

Here is what is confusing me:

---

Rewriting the given equation, we have:

. . . . .\(\displaystyle 2\, =\, \dfrac{\log_6(15x)}{\log_6(5x)}\)

. . . . .\(\displaystyle 2\, =\, \dfrac{\log_6(15)\, +\, \log_6(x)}{\log_6(5)\, +\, \log_6(x)}\)

Solving the equation for \(\displaystyle \log_6(x),\)

---

My question is : What's the rule that just allows you to rewrite the equation this way? I been working problems all day so maybe its just my mind is fried.

TIA.

There has always been such a rule since the dawn of abstraction. Essentially, whatever EQUAVALENT form you may be contemplating, simply substitute.

In your case, we have the long-standing rule of logarithms, \(\displaystyle \log(a\cdot b) = \log(a) + \log(b)\). This rule has not been around quite as long as abstraction or substitution, but still at least a couple hundred years.

Maybe I'm not understanding our question.

 

[doughishere]

Awww...thats what it is. I get it now...it combines both log(xy) and log(x/y) rules.



Mucho Gracias Sir.


Edit: Err...it doesnt combine them....but yeah i get it now.

 

[tkhunny]

log(x/y)

There is NONE of that. If you think there is, please show your work so that we can help you on the right path.

 

[doughishere]

There is NONE of that. If you think there is, please show your work so that we can help you on the right path.

No there is none of that......i think thats why I got confused..i knew that the division wasnt right but still got hooked on the division......its so strange ive literally been doing those problems all day. And this was like problem number 100ish for the day....i think im tired...time to call it a night. Thanks.


Edit: the really strange thing is for each problem i do i "more often than not" write down the rule...ie log[a](xy) = log[a](x) + log[a](y) and the division property which is log[a](x) / log[a](y) = log[a](x/y) and then solve the problem.

 

[tkhunny]

and the division property which is log[a](x) / log[a](y) = log[a](x/y)

This is where I thought you were going. There is NO such rule. Logarithms don't work that way.

\(\displaystyle \dfrac{\log_{a}(x)}{\log_{a}(y)} = \log_{y}(x)\)

\(\displaystyle \log_{a}(x/y) = \log_{a}(x) - \log_{a}(y)\)

Be more careful.

 

[doughishere]

Still sleeping on the job Doug?

Honestly. Dont tell the kids but I get really high some times and do my problems. Its not all the time but its really interesting to see how my mind kind of thinks and kind of put it all together.

That and yes I do have a mental laziness thats difficult for me. Lets just call it what it is. Anyways. Back to work. Doing every problem at the end of the chapters.

Thanks for the help all! Merry Christmas! Happy Holidays!


Mental Note: Be More Careful.